Positions of Large Groups LT830

In a string S of lowercase letters, these letters form consecutive groups of the same character.

For example, a string like S = "abbxxxxzyy" has the groups "a", "bb", "xxxx", "z" and "yy".

Call a group large if it has 3 or more characters.  We would like the starting and ending positions of every large group.

The final answer should be in lexicographic order.

 

Example 1:

Input: "abbxxxxzzy"
Output: [[3,6]]
Explanation: "xxxx" is the single large group with starting  3 and ending positions 6.

Example 2:

Input: "abc"
Output: []
Explanation: We have "a","b" and "c" but no large group.

Example 3:

Input: "abcdddeeeeaabbbcd"
Output: [[3,5],[6,9],[12,14]]

 

Note:  1 <= S.length <= 1000

Idea 1. two pointer, extend to the right until new element appears or reach the end of the input array.

Time complexity: O(n)

Space complexity: O(1) if not considering output List

loop the left pointer, as left pointer is fixed, expand the right pointer

class Solution {
    public List<List<Integer>> largeGroupPositions(String S) {
        List<List<Integer>> result = new ArrayList<>();
        for(int i = 0, right = 0; i < S.length(); i = right) {
          
            while(right < S.length() && S.charAt(right) == S.charAt(i)) {
              ++right;
            }
          
            if(right - i >= 3) {   
              result.add(Arrays.asList(i, right-1));
            } 
        }
        return result;
    }
}

Idea 1.a loop right pointer instead

class Solution {
    public List<List<Integer>> largeGroupPositions(String S) {
        List<List<Integer>> result = new ArrayList<>();
        for(int i = 0, right = 0; right < S.length(); ++right) {
          
            while(right+1 < S.length() && S.charAt(right) == S.charAt(right+1)) {
              ++right;
            }
          
            if(right - i + 1 >= 3) {   
              result.add(Arrays.asList(i, right));
            } 
            i = right + 1;
        }
        return result;
    }
}